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For the reaction: \(\mathrm{aA}+\mathrm{bB} \rightarrow \mathrm{P}\), \(r=K[\mathrm{~A}]^{a} \cdot[\mathrm{B}]^{b} .\) If concentration of \(\mathrm{A}\) is doubled, the rate is doubled. If concentration of \(\mathrm{B}\) is doubled, the rate becomes four times. The correct relation is (a) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (b) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+2 \cdot \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (c) \(-2 \cdot \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (d) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\)

Short Answer

Expert verified
(b) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+2 \cdot \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\)

Step by step solution

01

Identify the rate orders

From the given rate equation, identify the reaction orders with respect to A and B by analyzing the effect on the rate when their concentrations are changed. If doubling the concentration of A results in a doubling of the rate, it suggests that the reaction is first order with respect to A, or a = 1. Similarly, if doubling the concentration of B results in the rate becoming four times, it suggests that the reaction is second order with respect to B, or b = 2.
02

Write the rate equation with determined orders

Write the rate equation with the determined reaction orders. The rate equation is now known to be: \(r = K[\mathrm{A}]^{1} \cdot [\mathrm{B}]^{2}\).
03

Determine the relation between the rates of depletion

Establish the relationship between the rate of depletion of A and B. The stoichiometric coefficients (a and b) in the balanced chemical equation dictate the ratio in which reactants are consumed. Since the rate equation depends on both the stoichiometry of the reaction and the reaction orders, if a = 1 and b = 2, for every one mole of A that reacts, two moles of B will react. Therefore, the correct relation is \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t} = 2 \cdot \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) because for every decrement in A, there is twice the decrement in B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Order
Understanding reaction order is crucial when studying the dynamics of a chemical reaction. It indicates how the rate of reaction is affected by the concentration of each reactant. If the rate doubles when the concentration of a reactant is doubled, this is a typical behavior of a first-order reaction with respect to that reactant. In our exercise, when the concentration of substance A is doubled and the reaction rate also doubles, it confirms that A influences the rate first-order dependence, which is mathematically represented as 'a=1'.

For reactant B, quadrupling the rate upon doubling the concentration reveals a second-order relationship (the rate is proportional to the concentration squared), expressed as 'b=2'. This insight is foundational for writing the rate law equation, which quantifies the relationship between reactant concentration and the speed of the reaction.
Rate Law
The rate law is a mathematical equation that defines the relationship between the rate of a chemical reaction and the concentration of its reactants. This expression is vital because it allows chemists to predict how changes in conditions will affect the reaction speed. From the given exercise, with the identified reaction orders, we can express the rate law as: \(r = K[\mathrm{A}]^{1} \cdot [\mathrm{B}]^{2}\), where 'K' is the rate constant, a measure of the speed at which the reaction occurs.

This relationship is not necessarily apparent from the chemical equation itself but must be determined experimentally. Knowledge of the rate law helps in planning industrial chemical processes and understanding natural phenomena.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantification of the participants in a chemical reaction. It involves the study of the relative quantities of reactants and products in reactions and is governed by the balanced chemical equation. Stoichiometry is rooted in the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

In this exercise, stoichiometry tells us the mole ratio in which reactants combine and products form. The coefficients 'a' and 'b' serve as a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. They guide us to understand how much of substance B is needed to react with substance A and how their rates of depletion are related.
Chemical Kinetics
Chemical kinetics is the study of the speed or rate at which chemical reactions occur, as well as the factors that influence this speed. It includes not only the analysis of how fast reactions proceed but also an understanding of the steps that take place from reactants to products—the reaction mechanism.

The rate equation from our exercise provides a snapshot of kinetics in action. By examining the rates at which reactants are consumed or products are formed, chemists can deduce the order of the reaction and its rate law. Factors like temperature, catalysts, and the physical state of the reactants can significantly affect reaction rates. Chemical kinetics is indispensable in various fields, from designing pharmaceuticals to manufacturing materials and environmental science.

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Most popular questions from this chapter

The reaction: \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{NO}_{2}+\mathrm{OH}^{-}\) \(\rightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}^{-}-\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) obeys the rate law for pseudo first-order kinetics in the presence of a large excess of hydroxide ion. If \(1 \%\) of nitro ethane undergoes reaction in half a minute when the reactant concentration is \(0.002 \mathrm{M}\), what is the pseudo first-order rate constant? (a) \(2 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(6 \times 10^{-3} \mathrm{~min}^{-1}\) (c) \(4 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(1 \times 10^{-2} \mathrm{~min}^{-1}\)

For the second-order reaction: \(2 \mathrm{~A} \rightarrow \mathrm{B}\), time taken for the \([\mathrm{A}]\) to fall to one-fourth value is how many times the time it takes for \([\mathrm{A}]\) to fall to half of its initial value? (a) two (b) three (c) four (d) seven

If the rate of a gaseous reaction is independent of partial pressure of reactant, the order of reaction is (a) 0 (b) 1 (c) 2 (d) 3

The decomposition of hydrogen peroxide in aqueous solution is a first-order reaction: Time in min \(0 \quad 10\) Volume \((V\), in \(\mathrm{ml}) \quad 25.0 \quad 20.0\) where \(V\) is the number of \(\mathrm{ml}\) of potassium permanganate required to decompose a definite volume of hydrogen peroxide solution, at the given time. From the following data, the rate constant (in \(\min ^{-1}\) ) of reaction is \((\ln 5=1.6, \ln 2=0.7)\) (a) \(0.09\) (b) \(0.02\) (c) \(0.2\) (d) \(0.16\)

\(\begin{array}{lll}\text { For the } & \text { sequential } & \text { reactions: }\end{array}\) \(\mathrm{A} \stackrel{K_{1}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{C}\), the initial concentration of 'A' was \(0.2 \mathrm{M}\) and initially 'B' and 'C' were absent. The time at which the concentration of ' \(\mathrm{B}\) ' becomes maximum and the maximum concentration of ' B' are, respectively, (a) \(50 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\) (b) \(50 \mathrm{~min}, 0.2 \mathrm{M}\) (c) infinite, \(0.2 \mathrm{M}\) (d) \(25 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\)

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